1. Chapter 7 Class 12 Integrals
  2. Serial order wise
  3. Miscellaneous
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Question 4 (MCQ) - Miscellaneous - Chapter 7 Class 12 Integrals

Last updated at Dec. 16, 2024 by Teachoo

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Misc 44 The value of ∫_0^1▒tan^(−1)⁡((2𝑥 − 1)/(1 + 𝑥 − 𝑥^2 )) 𝑑𝑥 is equal to (A) 1 (B) 0 (C) −1 (D) 𝜋/4 Let I=∫_0^1▒tan^(−1)⁡((2𝑥−1)/(1 + 𝑥 − 𝑥^2 )) 𝑑𝑥 I=∫_0^1▒tan^(−1)⁡[(𝑥 + (𝑥 − 1))/(1 + 𝑥(1 − 𝑥) )] 𝑑𝑥 I=∫_0^1▒tan^(−1)⁡[(𝑥 + (𝑥 − 1))/(1 + 𝑥(𝑥 − 1) )] 𝑑𝑥 Using 〖𝑡𝑎𝑛〗^(−1) 𝑥+〖𝑡𝑎𝑛〗^(−1) (𝑦)=〖𝑡𝑎𝑛〗^(−1)⁡[(𝑥 + 𝑦)/(1 − 𝑥𝑦)] ∴ I=∫_0^1▒[tan^(−1) (𝑥)+tan^(−1) (𝑥−1)] 𝑑𝑥 I=∫_0^1▒〖tan^(−1) [−(𝑥−1)] 〗 𝑑𝑥+∫_0^1▒〖tan^(−1) (−𝑥) 〗 𝑑𝑥 Using The Property, P4 : ∫_0^𝑎▒〖𝑓(𝑥)𝑑𝑥=〗 ∫_0^𝑎▒𝑓(𝑎−𝑥)𝑑𝑥 ∴ I=∫_0^1▒〖tan^(−1) (1−𝑥) 〗 𝑑𝑥+∫_0^1▒〖tan^(−1) (1−𝑥−1) 〗 𝑑𝑥 I=∫_0^1▒〖tan^(−1) [−(𝑥−1)] 〗 𝑑𝑥+∫_0^1▒〖tan^(−1) (−𝑥) 〗 𝑑𝑥 I=−∫_0^1▒〖tan^(−1) (𝑥−1) 〗 𝑑𝑥−∫_0^1▒〖tan^(−1) (𝑥) 〗 𝑑𝑥 tan^(−1) (−𝑥)=−tan^(−1)⁡〖(𝑥)〗 Adding (1) and (2) I+I =∫_0^1▒〖tan^(−1) (𝑥) 〗 𝑑𝑥+∫_0^1▒〖tan^(−1) (𝑥−1) 〗 𝑑𝑥−∫_0^1▒〖tan^(−1) (𝑥) 〗 𝑑𝑥−∫_0^1▒〖tan^(−1) (𝑥−1) 〗 𝑑𝑥 2I=0 ∴ I=0 Hence, correct answer is B.
  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

    3. Miscellaneous

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Case Based Questions (MCQ) →
Chapter 7 Class 12 Integrals
Serial order wise

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo